First of all, you need to prove that $\left\{s_n\right\}_{n \in \mathbb{N}} = \left\{ \frac{1}{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+n}} \right\}_{n \in \mathbb{N}} $ converges. In order to do that, we will find $\left\{a_n\right\}_{n \in \mathbb{N}} $ and $\left\{b_n\right\}_{n \in \mathbb{N}}$ such that:

$a_n \leq s_n \leq b_n$ for all $n \in \mathbb{N}$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = l$

And in this situation, we will be able to conclude that $\lim_{n \to \infty} s_n = l$.

Since $\frac{1}{\sqrt{n^2+n}} \leq \frac{1}{\sqrt{n^2+i}} \leq \frac{1}{\sqrt{n^2+1}}$ for all $i\in\left\{1, \ldots, n \right\}$, we have that $\frac{n}{\sqrt{n^2+n}} \leq s_n \leq \frac{n}{\sqrt{n^2+1}}$.

Moreover, $\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} = 1$. That can be easily seen dividing numerator and denominator by $n$ in both sequences.